Using tangent substitution on $Bbb R$












0












$begingroup$


How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.






calculus integration







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$endgroup$








  • 1




    $begingroup$
    Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 21:20










  • $begingroup$
    On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
    $endgroup$
    – Gaboru
    Dec 2 '18 at 21:41










  • $begingroup$
    The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 22:15


















0












$begingroup$


How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.






calculus integration







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 21:20










  • $begingroup$
    On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
    $endgroup$
    – Gaboru
    Dec 2 '18 at 21:41










  • $begingroup$
    The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 22:15
















0












0








0





$begingroup$


How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.






calculus integration







share|cite|improve this question











$endgroup$




How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 23:09









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Dec 2 '18 at 20:09









GaboruGaboru

637




637








  • 1




    $begingroup$
    Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 21:20










  • $begingroup$
    On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
    $endgroup$
    – Gaboru
    Dec 2 '18 at 21:41










  • $begingroup$
    The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 22:15
















  • 1




    $begingroup$
    Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 21:20










  • $begingroup$
    On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
    $endgroup$
    – Gaboru
    Dec 2 '18 at 21:41










  • $begingroup$
    The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
    $endgroup$
    – Jean Marie
    Dec 2 '18 at 22:15










1




1




$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20




$begingroup$
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
$endgroup$
– Jean Marie
Dec 2 '18 at 21:20












$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41




$begingroup$
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
$endgroup$
– Gaboru
Dec 2 '18 at 21:41












$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15






$begingroup$
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
$endgroup$
– Jean Marie
Dec 2 '18 at 22:15












1 Answer
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$begingroup$

Have a look at the discontinuous curve associated with your function :



enter image description here



It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.






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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Have a look at the discontinuous curve associated with your function :



    enter image description here



    It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Have a look at the discontinuous curve associated with your function :



      enter image description here



      It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Have a look at the discontinuous curve associated with your function :



        enter image description here



        It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.






        share|cite|improve this answer











        $endgroup$



        Have a look at the discontinuous curve associated with your function :



        enter image description here



        It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, on each $D_k$ with a specific $C_k$, independent of the others.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 19:45

























        answered Dec 2 '18 at 23:07









        Jean MarieJean Marie

        29.3k42050




        29.3k42050






























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